The Schrodinger wave equation for hydrogen atom is Ps ("radial") =1/(16sqrt4) ((Z)/(a_0))^(3//2) [ (sigma -1) (sigma^2 - 8 sigma + 12 )]e^(-sigma//2) where a_0 and z are the constant in which answer can be expressed and sigma =(2pi)/(a_0) minimum and maximum position of radial nodes from nucleus are

The Schrodinger wave equation for hydrogen atom is Ps ("radial") =1/(1

1.	The Schrodinger wave equation for hydrogen atom is Ps ("radial") =1/(16sqrt4) ((Z)/(a_0))^(3//2) [ (sigma -1) (sigma^2 - 8 sigma + 12 )]e^(-sigma//2) where a_0 and z are the constant in which answer can be expressed and sigma =(2pi)/(a_0) minimum and maximum position of radial nodes from nucleus are
Answer» `(a_0)/(Z), (3a_0)/(Z)` `(a_0)/(2Z) ,(a_0)/(Z)` `(a_0)/(2Z), (3a_0)/(Z)` `(a_0)/(2Z), (4a_0)/(Z)` Solution :Probability of FINDING `e^-` is zero impliesthat `Psi^2 = 0 ` or `Psi=0` `IMPLIES(sigma -1) = 0 impliessigma =1` or `r_1 =(a_0)/(2z) ` or `(sigma^2 - 8 sigma + 12) =0 ` and `(sigma-6 ) (sigma-2)=0` `sigma=6, r= (3a_0)/(z) , sigma = 2 , r = (a_0)/(z) , r_2 = (3a_0)/(z)`